How do I get the power series for $\frac{z^2}{(1+z)^2}$

Question

How do I get the power series for $\frac{z^2}{(1+z)^2}$ around $z=0$? I know the answer is: $\sum_{n=1}(-1)^{n+1}nz^{n+1}$, for $|z|<1$ and $\sum_{n=1}(-1)^{n+1}nz^{-n+2}$, for $|z|>1$. I've tried using the geometric series and working the expressions but getting the $n$ before the $z^n$ is impossible doing that. So I have no idea on how to do this. Any help?

Answer 1

It is well-known that the power series of $f(z)=\dfrac{1}{1+z}$ is:

$$f(z)=\dfrac{1}{1+z}=\dfrac{1}{1-(-z)}=\sum_{n=0}^\infty (-z)^n=\sum_{n=0}^\infty (-1)^nz^n.$$ By taking the derivative we obtain

$$f'(z)=\dfrac{-1}{(1+z)^2}=\sum_{n=1}^\infty (-1)^n nz^{n-1}$$

and finally, we multiply by $-z^2$ in both sides to get

$$\begin{align*}\dfrac{z^2}{(1+z)^2}&=(-z^2)\cdot \sum_{n=1}^\infty (-1)^n\cdot n\cdot z^{n-1}\\ &=\sum_{n=1}^\infty (-1)^{n+1}\cdot n\cdot z^{(n-1)+2}\\ &=\sum_{n=1}^\infty (-1)^{n+1}\cdot n\cdot z^{n+1} \end{align*}$$